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Save portbury/41a02fe35b48e612857df847a7e1fe97 to your computer and use it in GitHub Desktop.
$xeroPayrollAuApi = new XeroAPI\XeroPHP\Api\PayrollAuApi( | |
new GuzzleHttp\Client(), | |
$xeroApiConnection["config"] | |
); | |
$xero_id = "00000000-aaaa-0000-aaaa-000000000000"; | |
$homeemail = "mark@gmail.com"; | |
$mobile = "0427 123 456"; | |
$employee = new XeroAPI\XeroPHP\Models\PayrollAu\Employee; | |
$employee->setEmail($homeemail); | |
$employee->setMobile($mobile); | |
try { | |
$resultPayroll = $xeroPayrollAuApi->updateEmployee($xeroApiConnection["tenantId"], $xero_id, $employee); | |
} catch (Exception $e) { | |
echo $e->getMessage(); | |
} |
any updates? i am having a similar problem:
System.Runtime.Serialization.SerializationException: Type definitions should start with a '{', expecting serialized type 'ApiException', got string starting with: oauth_problem=signature_method_rejected&oauth_prob at Xero.Api.Infrastructure.ThirdParty.ServiceStack.Text.Common.DeserializeTypeRefJson.StringToType(Type type, String strType, EmptyCtorDelegate ctorFn, Dictionary2 typeAccessorMap) at Xero.Api.Infrastructure.ThirdParty.ServiceStack.Text.JsonSerializer.DeserializeFromString[T](String value) at Xero.Api.Infrastructure.Http.XeroHttpClient.HandleErrors(Response response) at Xero.Api.Infrastructure.Http.XeroHttpClient.Read[TResult,TResponse](Response response) at Xero.Api.Common.XeroReadEndpoint
3.Get(String endpoint, String child) at Xero.Api.Core.XeroCoreApi.get_Organisation()
Error being received back when dumping $resultPayroll:
Bad requestError occurred during JSON de/serialization. Cannot deserialize the current JSON object (e.g. {"name":"value"}) into type 'Paycycle.API.DTO.AU.Employee.UpdateEmployeeRequest' because the type requires a JSON array (e.g. [1,2,3]) to deserialize correctly. To fix this error either change the JSON to a JSON array (e.g. [1,2,3]) or change the deserialized type so that it is a normal .NET type (e.g. not a primitive type like integer, not a collection type like an array or List<T>) that can be deserialized from a JSON object. JsonObjectAttribute can also be added to the type to force it to deserialize from a JSON object. Path 'Email', line 1, position 13."